Question

A steam power plant operates on an ideal Rankine cycle with two of its feedwater heater closed as shown in the figure below and with a net power output of 105MW. The mass flow rate of the steam is 70kg/s and enters the high-pressure turbine at 6MPa and $740^\circ C$ and leaves at 1.5 Mpa. a fraction of x of steam is extracted at this pressure to the closed feedwater. a fraction of y of steam is extracted from the second stage of the turbine at 250 kPa and is also fed into the second closed feedwater heater that operates at 250 kPa. The rest of the steam then expanded in the low-pressure turbine to the condenser pressure of 20kPa. the condensate exiting both of the feedwater heaters is saturated liquid. Assume ideal state for the feedwater that exits from both of the closed feedwater heaters. Assume the working fluid experiences no irreversibilities as it passes through the turbine, pump, boiler, and condenser.

a). Draw the T-s diagram of the ideal cycle

b). Complete the table below

c). determine the fraction of steam, x that is extracted from the turbine at point 6

d). determine the fraction of steam, y that is extracted from the turbine at point 7

e). determine the thermal efficiency

Detailed information:

State 5: P = 6MPa, T = $740^\circ C$ , S = 7.5190 kJ/kg.K

state	enthalpy, h (kJ/kg)
1
2
3
4
5	3989
6	3406
7	2918
8
9
10
11
12

Figure:

Answer 1

a. T-S diagram is as follows

5 6 MPa T 6 q 1.5 MPa 1 Ч 0.25 MPa 7 10 0:02 mpa 12 8 > Pace No. s

b. There is no information about Point 3 and point 4. So it's not possible to evaluate the Enthalpies of these two points. Rest all are as follows.

1 &2
P= 0.02 MPa h = 251.42 Si= 0.83202 20 Ps GMPa, S2=51 0.83202 h₂= 256.20 2
8,9,10,11,12
8 5,= 0.83202 ss = 7.5190 P8 = 0.02 MPa S8=55 Ses st x8 (Sfg78 7.5190- 0.83202 + 28 (7.90720-0.83202) 0.9451 > x8 = ho: hit x8 (htg); ho= 251.42 t 0.9451 (2608.9 – 251.42) ho a 2479.47 q To hosha = 844.56 Ps los mpa has 844.56 11 P = 0.25 mpa Thu = 535.34

h12 = h11 = 535.34

C&D After taking a proper assumption h3 h4 can be calculated and based on that x & y can be calculated using the following equations:

3) &14 2-> 3 -> & (hu-hz )= x (ho-ha) and (ha-ha 2 = y Chy-hil tachio-hu